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Types of Forces>Work Done by a Variable Force

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Topics

  • Work Done by Variable Forces: The Power of Integration
  • Dividing and Conquering (Integration)
  • The Graphical Method: Area Under the Curve
  • Example  
Maharashtra State Board: Class 11

Work Done by a Variable Force: The Power of Integration

The formula for work you already know is:

W = F . s = F s cos θ

Crucial Limitation: This formula ONLY works when the force (F) is constant throughout the displacement (s).

Real-World Check: In many real situations, the force changes! We call this a Variable Force.

  • If you launch a satellite very far from Earth, the gravitational force pulling it changes as its distance from Earth increases.
  • When you stretch a spring, the force required increases the more you stretch it. This is a classic variable force!
Maharashtra State Board: Class 11

Dividing and Conquering (Integration)

Since the force F is constantly changing, we can't use W = F × Δs because we don't know which value of F to use.

The brilliant method is to break the total displacement (s1 to s2) into a large number of infinitely tiny segments, called ds.

  1. Focus on a Tiny Segment (ds): This segment is so incredibly small that the force F over this distance is practically constant.

  2. Work for the Segment (dW): For this tiny segment, the work done (dW) is simply:
    Work = Force × Tiny Displacement ⇒ dW = F . ds

  3. Total Work (W): To find the total work, we just add up (integrate) all the tiny dW's from the start point (s1​) to the end point (s2​).
    W = \[\sum dW=\int_{s_{1}}^{s_{2}}F\cdot ds\]

Maharashtra State Board: Class 11

The Graphical Method: Area Under the Curve

The total work done by a variable force is equal to the area under the Force-Displacement (F-s) graph between the initial and final displacements.

  • The area of any curve segment is calculated by Height × Width. In our F-s graph, the area of a tiny strip is F × ds, which is exactly our definition of the tiny work dW. Summing these strips (integration) gives the total area, which is the total work, W.

  • Linear Variation (Easy Case): If the force varies linearly (a straight line, like a spring), the area is a simple geometric shape (a trapezium or triangle). See Fig 4.1(b).

Fig 4.1 (b): Work done by linearly varying force.

  • Non-linear Variation (Calculus Case): If the force varies non-linearly (a curve), you must use integration (calculus) to find the area. See Fig. 4.1(a).

Fig 4.1 (a): Work done by a nonlinearly varying force. 

Maharashtra State Board: Class 11

Example

Example 4.4: Over a given region, a force (in newtons) varies as F = 3x2 - 2x + 1. An object is displaced from x1 = 20 cm to x2 = 40 cm by the given force. Calculate the amount of work done.

Solution:

Step 1: Check Units and Convert to SI (Crucial!):
The Force equation is in Newtons (SI unit), so Displacement must be in Meters (SI unit).
x1 = 20 cm = 0.2 m
x2 = 40 cm = 0.4 m

Step 2: Apply the Integration Formula:
W = \[\int_{x_1}^{x_2}F\cdot dx=\int_{0.2}^{0.4}(3x^2-2x+1)dx\]

Step 3: Perform the Integration:
W = \[\left[\frac{3x^3}{3}-\frac{2x^2}{2}+x\right]_{0.2}^{0.4}=[x^3-x^2+x]_{0.2}^{0.4}\]

Step 4: Substitute the Limits and Calculate:
W = [(0.4)3 - (0.4)2 + 0.4] - [(0.2)3 - (0.2)2 + 0.2]
W = [0.064 - 0.16 + 0.4] - [0.008 - 0.04 + 0.2]
W = [0.304] - [0.168]
W = 0.136 J

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